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Prove It Assuming you meant $\displaystyle \displaystyle \begin{align*} 2 - \frac{1}{a + 2} = \frac{a + 18}{4a^2 - 16} \end{align*}$, which it could very well not be because you didn't use brackets like you were asked to... First note that $\displaystyle \displaystyle \begin{align*} a \neq \pm 2 \end{align*}$ (why?), then
$\displaystyle \displaystyle \begin{align*} 2 - \frac{1}{a + 2} &= \frac{a + 18}{4a^2 - 16} \\ 2 - \frac{1}{a + 2} &= \frac{a + 18}{4(a - 2)(a + 2)} \\ \frac{2\cdot 4(a - 2)(a + 2)}{4(a - 2)(a + 2)} - \frac{4(a - 2)}{4(a - 2)(a + 2)} &= \frac{a + 18}{4(a - 2)(a + 2)} \\ \frac{8(a - 2)(a + 2) - 4(a - 2)}{4(a - 2)(a + 2)} &= \frac{a + 18}{4(a - 2)(a + 2)} \\ 8(a - 2)(a + 2) - 4(a - 2) &= a + 18 \end{align*}$
Can you continue?